Integrand size = 24, antiderivative size = 175 \[ \int \left (a+b x^2\right )^{3/2} \left (A+B x^2+C x^4\right ) \, dx=\frac {a \left (48 A b^2-a (8 b B-3 a C)\right ) x \sqrt {a+b x^2}}{128 b^2}+\frac {1}{192} \left (48 A-\frac {a (8 b B-3 a C)}{b^2}\right ) x \left (a+b x^2\right )^{3/2}+\frac {(8 b B-3 a C) x \left (a+b x^2\right )^{5/2}}{48 b^2}+\frac {C x^3 \left (a+b x^2\right )^{5/2}}{8 b}+\frac {a^2 \left (48 A b^2-a (8 b B-3 a C)\right ) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{128 b^{5/2}} \] Output:
1/128*a*(48*A*b^2-a*(8*B*b-3*C*a))*x*(b*x^2+a)^(1/2)/b^2+1/192*(48*A-a*(8* B*b-3*C*a)/b^2)*x*(b*x^2+a)^(3/2)+1/48*(8*B*b-3*C*a)*x*(b*x^2+a)^(5/2)/b^2 +1/8*C*x^3*(b*x^2+a)^(5/2)/b+1/128*a^2*(48*A*b^2-a*(8*B*b-3*C*a))*arctanh( b^(1/2)*x/(b*x^2+a)^(1/2))/b^(5/2)
Time = 0.23 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.80 \[ \int \left (a+b x^2\right )^{3/2} \left (A+B x^2+C x^4\right ) \, dx=\frac {\sqrt {b} x \sqrt {a+b x^2} \left (-9 a^3 C+6 a^2 b \left (4 B+C x^2\right )+16 b^3 x^2 \left (6 A+4 B x^2+3 C x^4\right )+8 a b^2 \left (30 A+14 B x^2+9 C x^4\right )\right )-3 a^2 \left (48 A b^2+a (-8 b B+3 a C)\right ) \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{384 b^{5/2}} \] Input:
Integrate[(a + b*x^2)^(3/2)*(A + B*x^2 + C*x^4),x]
Output:
(Sqrt[b]*x*Sqrt[a + b*x^2]*(-9*a^3*C + 6*a^2*b*(4*B + C*x^2) + 16*b^3*x^2* (6*A + 4*B*x^2 + 3*C*x^4) + 8*a*b^2*(30*A + 14*B*x^2 + 9*C*x^4)) - 3*a^2*( 48*A*b^2 + a*(-8*b*B + 3*a*C))*Log[-(Sqrt[b]*x) + Sqrt[a + b*x^2]])/(384*b ^(5/2))
Time = 0.26 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.87, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1473, 299, 211, 211, 224, 219}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \left (a+b x^2\right )^{3/2} \left (A+B x^2+C x^4\right ) \, dx\) |
| \(\Big \downarrow \) 1473 |
| \(\displaystyle \frac {\int \left (b x^2+a\right )^{3/2} \left ((8 b B-3 a C) x^2+8 A b\right )dx}{8 b}+\frac {C x^3 \left (a+b x^2\right )^{5/2}}{8 b}\) |
| \(\Big \downarrow \) 299 |
| \(\displaystyle \frac {\frac {\left (48 A b^2-a (8 b B-3 a C)\right ) \int \left (b x^2+a\right )^{3/2}dx}{6 b}+\frac {x \left (a+b x^2\right )^{5/2} (8 b B-3 a C)}{6 b}}{8 b}+\frac {C x^3 \left (a+b x^2\right )^{5/2}}{8 b}\) |
| \(\Big \downarrow \) 211 |
| \(\displaystyle \frac {\frac {\left (48 A b^2-a (8 b B-3 a C)\right ) \left (\frac {3}{4} a \int \sqrt {b x^2+a}dx+\frac {1}{4} x \left (a+b x^2\right )^{3/2}\right )}{6 b}+\frac {x \left (a+b x^2\right )^{5/2} (8 b B-3 a C)}{6 b}}{8 b}+\frac {C x^3 \left (a+b x^2\right )^{5/2}}{8 b}\) |
| \(\Big \downarrow \) 211 |
| \(\displaystyle \frac {\frac {\left (48 A b^2-a (8 b B-3 a C)\right ) \left (\frac {3}{4} a \left (\frac {1}{2} a \int \frac {1}{\sqrt {b x^2+a}}dx+\frac {1}{2} x \sqrt {a+b x^2}\right )+\frac {1}{4} x \left (a+b x^2\right )^{3/2}\right )}{6 b}+\frac {x \left (a+b x^2\right )^{5/2} (8 b B-3 a C)}{6 b}}{8 b}+\frac {C x^3 \left (a+b x^2\right )^{5/2}}{8 b}\) |
| \(\Big \downarrow \) 224 |
| \(\displaystyle \frac {\frac {\left (48 A b^2-a (8 b B-3 a C)\right ) \left (\frac {3}{4} a \left (\frac {1}{2} a \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}+\frac {1}{2} x \sqrt {a+b x^2}\right )+\frac {1}{4} x \left (a+b x^2\right )^{3/2}\right )}{6 b}+\frac {x \left (a+b x^2\right )^{5/2} (8 b B-3 a C)}{6 b}}{8 b}+\frac {C x^3 \left (a+b x^2\right )^{5/2}}{8 b}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {\left (\frac {3}{4} a \left (\frac {a \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 \sqrt {b}}+\frac {1}{2} x \sqrt {a+b x^2}\right )+\frac {1}{4} x \left (a+b x^2\right )^{3/2}\right ) \left (48 A b^2-a (8 b B-3 a C)\right )}{6 b}+\frac {x \left (a+b x^2\right )^{5/2} (8 b B-3 a C)}{6 b}}{8 b}+\frac {C x^3 \left (a+b x^2\right )^{5/2}}{8 b}\) |
Input:
Int[(a + b*x^2)^(3/2)*(A + B*x^2 + C*x^4),x]
Output:
(C*x^3*(a + b*x^2)^(5/2))/(8*b) + (((8*b*B - 3*a*C)*x*(a + b*x^2)^(5/2))/( 6*b) + ((48*A*b^2 - a*(8*b*B - 3*a*C))*((x*(a + b*x^2)^(3/2))/4 + (3*a*((x *Sqrt[a + b*x^2])/2 + (a*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(2*Sqrt[b]) ))/4))/(6*b))/(8*b)
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 )), x] + Simp[2*a*(p/(2*p + 1)) Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ {a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] && !GtQ[a, 0]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[d*x *((a + b*x^2)^(p + 1)/(b*(2*p + 3))), x] - Simp[(a*d - b*c*(2*p + 3))/(b*(2 *p + 3)) Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && NeQ[2*p + 3, 0]
Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Simp[c^p*x^(4*p - 1)*((d + e*x^2)^(q + 1)/(e*(4*p + 2*q + 1))) , x] + Simp[1/(e*(4*p + 2*q + 1)) Int[(d + e*x^2)^q*ExpandToSum[e*(4*p + 2*q + 1)*(a + b*x^2 + c*x^4)^p - d*c^p*(4*p - 1)*x^(4*p - 2) - e*c^p*(4*p + 2*q + 1)*x^(4*p), x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && !LtQ[q, -1]
Time = 0.56 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.71
| method | result | size |
| pseudoelliptic | \(\frac {\frac {3 \left (b^{2} A -\frac {1}{6} a b B +\frac {1}{16} a^{2} C \right ) a^{2} \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{x \sqrt {b}}\right )}{8}+\frac {5 \sqrt {b \,x^{2}+a}\, \left (a \left (\frac {3}{10} C \,x^{4}+\frac {7}{15} x^{2} B +A \right ) b^{\frac {5}{2}}+\frac {2 \left (\frac {1}{2} C \,x^{4}+\frac {2}{3} x^{2} B +A \right ) x^{2} b^{\frac {7}{2}}}{5}+\frac {\left (\left (\frac {C \,x^{2}}{4}+B \right ) b^{\frac {3}{2}}-\frac {3 C a \sqrt {b}}{8}\right ) a^{2}}{10}\right ) x}{8}}{b^{\frac {5}{2}}}\) | \(124\) |
| risch | \(\frac {x \left (48 b^{3} C \,x^{6}+64 B \,b^{3} x^{4}+72 C a \,b^{2} x^{4}+96 A \,x^{2} b^{3}+112 B \,x^{2} a \,b^{2}+6 C \,a^{2} b \,x^{2}+240 a \,b^{2} A +24 a^{2} b B -9 C \,a^{3}\right ) \sqrt {b \,x^{2}+a}}{384 b^{2}}+\frac {a^{2} \left (48 b^{2} A -8 a b B +3 a^{2} C \right ) \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{128 b^{\frac {5}{2}}}\) | \(137\) |
| default | \(A \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )+C \left (\frac {x^{3} \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{8 b}-\frac {3 a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{6 b}-\frac {a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )}{6 b}\right )}{8 b}\right )+B \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{6 b}-\frac {a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )}{6 b}\right )\) | \(229\) |
Input:
int((b*x^2+a)^(3/2)*(C*x^4+B*x^2+A),x,method=_RETURNVERBOSE)
Output:
5/8/b^(5/2)*(3/5*(b^2*A-1/6*a*b*B+1/16*a^2*C)*a^2*arctanh((b*x^2+a)^(1/2)/ x/b^(1/2))+(b*x^2+a)^(1/2)*(a*(3/10*C*x^4+7/15*x^2*B+A)*b^(5/2)+2/5*(1/2*C *x^4+2/3*x^2*B+A)*x^2*b^(7/2)+1/10*((1/4*C*x^2+B)*b^(3/2)-3/8*C*a*b^(1/2)) *a^2)*x)
Time = 0.10 (sec) , antiderivative size = 304, normalized size of antiderivative = 1.74 \[ \int \left (a+b x^2\right )^{3/2} \left (A+B x^2+C x^4\right ) \, dx=\left [\frac {3 \, {\left (3 \, C a^{4} - 8 \, B a^{3} b + 48 \, A a^{2} b^{2}\right )} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (48 \, C b^{4} x^{7} + 8 \, {\left (9 \, C a b^{3} + 8 \, B b^{4}\right )} x^{5} + 2 \, {\left (3 \, C a^{2} b^{2} + 56 \, B a b^{3} + 48 \, A b^{4}\right )} x^{3} - 3 \, {\left (3 \, C a^{3} b - 8 \, B a^{2} b^{2} - 80 \, A a b^{3}\right )} x\right )} \sqrt {b x^{2} + a}}{768 \, b^{3}}, -\frac {3 \, {\left (3 \, C a^{4} - 8 \, B a^{3} b + 48 \, A a^{2} b^{2}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (48 \, C b^{4} x^{7} + 8 \, {\left (9 \, C a b^{3} + 8 \, B b^{4}\right )} x^{5} + 2 \, {\left (3 \, C a^{2} b^{2} + 56 \, B a b^{3} + 48 \, A b^{4}\right )} x^{3} - 3 \, {\left (3 \, C a^{3} b - 8 \, B a^{2} b^{2} - 80 \, A a b^{3}\right )} x\right )} \sqrt {b x^{2} + a}}{384 \, b^{3}}\right ] \] Input:
integrate((b*x^2+a)^(3/2)*(C*x^4+B*x^2+A),x, algorithm="fricas")
Output:
[1/768*(3*(3*C*a^4 - 8*B*a^3*b + 48*A*a^2*b^2)*sqrt(b)*log(-2*b*x^2 - 2*sq rt(b*x^2 + a)*sqrt(b)*x - a) + 2*(48*C*b^4*x^7 + 8*(9*C*a*b^3 + 8*B*b^4)*x ^5 + 2*(3*C*a^2*b^2 + 56*B*a*b^3 + 48*A*b^4)*x^3 - 3*(3*C*a^3*b - 8*B*a^2* b^2 - 80*A*a*b^3)*x)*sqrt(b*x^2 + a))/b^3, -1/384*(3*(3*C*a^4 - 8*B*a^3*b + 48*A*a^2*b^2)*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) - (48*C*b^4*x^ 7 + 8*(9*C*a*b^3 + 8*B*b^4)*x^5 + 2*(3*C*a^2*b^2 + 56*B*a*b^3 + 48*A*b^4)* x^3 - 3*(3*C*a^3*b - 8*B*a^2*b^2 - 80*A*a*b^3)*x)*sqrt(b*x^2 + a))/b^3]
Time = 0.43 (sec) , antiderivative size = 274, normalized size of antiderivative = 1.57 \[ \int \left (a+b x^2\right )^{3/2} \left (A+B x^2+C x^4\right ) \, dx=\begin {cases} \sqrt {a + b x^{2}} \left (\frac {C b x^{7}}{8} + \frac {x^{5} \left (B b^{2} + \frac {9 C a b}{8}\right )}{6 b} + \frac {x^{3} \left (A b^{2} + 2 B a b + C a^{2} - \frac {5 a \left (B b^{2} + \frac {9 C a b}{8}\right )}{6 b}\right )}{4 b} + \frac {x \left (2 A a b + B a^{2} - \frac {3 a \left (A b^{2} + 2 B a b + C a^{2} - \frac {5 a \left (B b^{2} + \frac {9 C a b}{8}\right )}{6 b}\right )}{4 b}\right )}{2 b}\right ) + \left (A a^{2} - \frac {a \left (2 A a b + B a^{2} - \frac {3 a \left (A b^{2} + 2 B a b + C a^{2} - \frac {5 a \left (B b^{2} + \frac {9 C a b}{8}\right )}{6 b}\right )}{4 b}\right )}{2 b}\right ) \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right ) & \text {for}\: b \neq 0 \\a^{\frac {3}{2}} \left (A x + \frac {B x^{3}}{3} + \frac {C x^{5}}{5}\right ) & \text {otherwise} \end {cases} \] Input:
integrate((b*x**2+a)**(3/2)*(C*x**4+B*x**2+A),x)
Output:
Piecewise((sqrt(a + b*x**2)*(C*b*x**7/8 + x**5*(B*b**2 + 9*C*a*b/8)/(6*b) + x**3*(A*b**2 + 2*B*a*b + C*a**2 - 5*a*(B*b**2 + 9*C*a*b/8)/(6*b))/(4*b) + x*(2*A*a*b + B*a**2 - 3*a*(A*b**2 + 2*B*a*b + C*a**2 - 5*a*(B*b**2 + 9*C *a*b/8)/(6*b))/(4*b))/(2*b)) + (A*a**2 - a*(2*A*a*b + B*a**2 - 3*a*(A*b**2 + 2*B*a*b + C*a**2 - 5*a*(B*b**2 + 9*C*a*b/8)/(6*b))/(4*b))/(2*b))*Piecew ise((log(2*sqrt(b)*sqrt(a + b*x**2) + 2*b*x)/sqrt(b), Ne(a, 0)), (x*log(x) /sqrt(b*x**2), True)), Ne(b, 0)), (a**(3/2)*(A*x + B*x**3/3 + C*x**5/5), T rue))
Time = 0.04 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.18 \[ \int \left (a+b x^2\right )^{3/2} \left (A+B x^2+C x^4\right ) \, dx=\frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} C x^{3}}{8 \, b} + \frac {1}{4} \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A x + \frac {3}{8} \, \sqrt {b x^{2} + a} A a x - \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} C a x}{16 \, b^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} C a^{2} x}{64 \, b^{2}} + \frac {3 \, \sqrt {b x^{2} + a} C a^{3} x}{128 \, b^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} B x}{6 \, b} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B a x}{24 \, b} - \frac {\sqrt {b x^{2} + a} B a^{2} x}{16 \, b} + \frac {3 \, C a^{4} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{128 \, b^{\frac {5}{2}}} - \frac {B a^{3} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{16 \, b^{\frac {3}{2}}} + \frac {3 \, A a^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {b}} \] Input:
integrate((b*x^2+a)^(3/2)*(C*x^4+B*x^2+A),x, algorithm="maxima")
Output:
1/8*(b*x^2 + a)^(5/2)*C*x^3/b + 1/4*(b*x^2 + a)^(3/2)*A*x + 3/8*sqrt(b*x^2 + a)*A*a*x - 1/16*(b*x^2 + a)^(5/2)*C*a*x/b^2 + 1/64*(b*x^2 + a)^(3/2)*C* a^2*x/b^2 + 3/128*sqrt(b*x^2 + a)*C*a^3*x/b^2 + 1/6*(b*x^2 + a)^(5/2)*B*x/ b - 1/24*(b*x^2 + a)^(3/2)*B*a*x/b - 1/16*sqrt(b*x^2 + a)*B*a^2*x/b + 3/12 8*C*a^4*arcsinh(b*x/sqrt(a*b))/b^(5/2) - 1/16*B*a^3*arcsinh(b*x/sqrt(a*b)) /b^(3/2) + 3/8*A*a^2*arcsinh(b*x/sqrt(a*b))/sqrt(b)
Time = 0.13 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.89 \[ \int \left (a+b x^2\right )^{3/2} \left (A+B x^2+C x^4\right ) \, dx=\frac {1}{384} \, {\left (2 \, {\left (4 \, {\left (6 \, C b x^{2} + \frac {9 \, C a b^{6} + 8 \, B b^{7}}{b^{6}}\right )} x^{2} + \frac {3 \, C a^{2} b^{5} + 56 \, B a b^{6} + 48 \, A b^{7}}{b^{6}}\right )} x^{2} - \frac {3 \, {\left (3 \, C a^{3} b^{4} - 8 \, B a^{2} b^{5} - 80 \, A a b^{6}\right )}}{b^{6}}\right )} \sqrt {b x^{2} + a} x - \frac {{\left (3 \, C a^{4} - 8 \, B a^{3} b + 48 \, A a^{2} b^{2}\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{128 \, b^{\frac {5}{2}}} \] Input:
integrate((b*x^2+a)^(3/2)*(C*x^4+B*x^2+A),x, algorithm="giac")
Output:
1/384*(2*(4*(6*C*b*x^2 + (9*C*a*b^6 + 8*B*b^7)/b^6)*x^2 + (3*C*a^2*b^5 + 5 6*B*a*b^6 + 48*A*b^7)/b^6)*x^2 - 3*(3*C*a^3*b^4 - 8*B*a^2*b^5 - 80*A*a*b^6 )/b^6)*sqrt(b*x^2 + a)*x - 1/128*(3*C*a^4 - 8*B*a^3*b + 48*A*a^2*b^2)*log( abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(5/2)
Timed out. \[ \int \left (a+b x^2\right )^{3/2} \left (A+B x^2+C x^4\right ) \, dx=\int {\left (b\,x^2+a\right )}^{3/2}\,\left (C\,x^4+B\,x^2+A\right ) \,d x \] Input:
int((a + b*x^2)^(3/2)*(A + B*x^2 + C*x^4),x)
Output:
int((a + b*x^2)^(3/2)*(A + B*x^2 + C*x^4), x)
Time = 0.16 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.05 \[ \int \left (a+b x^2\right )^{3/2} \left (A+B x^2+C x^4\right ) \, dx=\frac {-9 \sqrt {b \,x^{2}+a}\, a^{3} b c x +264 \sqrt {b \,x^{2}+a}\, a^{2} b^{3} x +6 \sqrt {b \,x^{2}+a}\, a^{2} b^{2} c \,x^{3}+208 \sqrt {b \,x^{2}+a}\, a \,b^{4} x^{3}+72 \sqrt {b \,x^{2}+a}\, a \,b^{3} c \,x^{5}+64 \sqrt {b \,x^{2}+a}\, b^{5} x^{5}+48 \sqrt {b \,x^{2}+a}\, b^{4} c \,x^{7}+9 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{4} c +120 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{3} b^{2}}{384 b^{3}} \] Input:
int((b*x^2+a)^(3/2)*(C*x^4+B*x^2+A),x)
Output:
( - 9*sqrt(a + b*x**2)*a**3*b*c*x + 264*sqrt(a + b*x**2)*a**2*b**3*x + 6*s qrt(a + b*x**2)*a**2*b**2*c*x**3 + 208*sqrt(a + b*x**2)*a*b**4*x**3 + 72*s qrt(a + b*x**2)*a*b**3*c*x**5 + 64*sqrt(a + b*x**2)*b**5*x**5 + 48*sqrt(a + b*x**2)*b**4*c*x**7 + 9*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt( a))*a**4*c + 120*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**3* b**2)/(384*b**3)